# extension Rolle's theorem for limit values

Rolle's theorem states that:

If a real-valued function $$f$$ is continuous on a proper closed interval $$[a, b]$$, differentiable on the open interval $$(a, b)$$, and $$f (a) = f (b)$$, then there exists at least one $$c$$ in the open interval $$(a, b)$$ such that $${\displaystyle f'(c)=0}$$.

Exercise:

Show that Rolle's theorem is true in case $$f$$ is defined and differentiable in the open interval ] $$a, b\left[\right.$$, and $$\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to b^-}f(x)$$. Note that $$a$$ could be $$-\infty$$ and $$b$$ could be $$+\infty$$. Furthermore the two limits could also be infinite.

My attempt:

Let choose $$a_1, b_1$$ so that $$a. Now we have the following cases:

$$f(a_1)=f(b_1)$$, $$f(a_1) or $$f(a_1)>f(b_1)$$.

When $$f(a_1)=f(b_1)$$ we can directly apply Rolle's theorem on $$]a_1,b_1[$$ so $$\exists c \in ]a_1,b_1[$$ such that $$f′(c)=0$$.

If $$f(a_1) then there is a real number $$z$$ such that $$f(a_1) and by the intermediate value theorem we have $$a_2 \in ]a_1,b_1[$$ which $$f(a_2)=z$$. Now, again we can apply Rolle's theorem to the restriction of $$f$$ to $$[a_2,b_1]$$.

The case when $$f(a_1)>f(b_1)$$ can be done similarly.

My question: Is this proof correct or I am missing something?

• You don't have $f(b_1)=z$ so you can't apply rolle like you did on $[a_2,b_1]$. Sep 15, 2021 at 18:24
• If those limits are finite just use a function $g$ with $g=f$ in $(a, b)$ and $g(a) =g(b)$ equals those limits. Sep 15, 2021 at 20:03
• Handle infinite limits separately. Sep 15, 2021 at 20:04
• I see. In that case, you need to follow the hints that @ParamanandSingh has just posted to reduce the problem to cases where you can apply Rolle's theorem. I think Paramanand should post the hints as an answer. Sep 15, 2021 at 20:18
• @Koro: check the first comment under this question. It gives the problem with the proof in question. Sep 16, 2021 at 15:59

As mentioned in comments, let us first handle the case when $$\lim_{x\to a^+} f(x) =\lim_{x\to b^-} f(x) \in\mathbb {R}$$ (meaning these limits are finite). We can define a function $$g:[a, b] \to\mathbb {R}$$ as $$g(x) =f(x) \, \forall x\in(a, b)$$ and $$g(a) =\lim_{x\to a^+} f(x), g(b) =\lim_{x\to b^-} f(x)$$ One can check that $$g'(x) =f'(x)$$ for all $$x\in(a, b)$$ and that $$g$$ satisfies all conditions for applicability of Rolle's theorem on $$[a, b]$$ and our job is done by applying Rolle's theorem on $$g$$.

Let me also say a few words about your approach. You can ensure that there are numbers $$a_1,b_1$$ in $$(a, b)$$ such that $$f(a_1)=f(b_1)$$ and there is no need to consider further cases.

The proof below is simple if you try to picture the graph of $$f$$ and use intermediate value theorem.

If $$f$$ is constant on $$(a, b)$$ then $$f'$$ is zero on whole of $$(a, b)$$. Otherwise $$f$$ takes at least two distinct values on $$(a, b)$$ and then one of these values must be different from the given equal limits $$\lim _{x\to a^+} f(x) =\lim _{x\to b^-} f(x) =L\, \text {(say)}$$ Thus we have a $$d\in(a, b)$$ with $$f(d) \neq L$$.

Let $$k$$ be a real number between $$f(d)$$ and $$L$$ (for clarity let $$L and the case $$L>k>f(d)$$ is similar). Choosing $$\epsilon =k-L>0$$ in definition of limit we can see that we have an interval of type $$(a, a+h_1)$$ where the values of $$f$$ are less than $$k$$ (write details explicitly using definition of limit as applied to $$\lim_{x\to a^+} f(x) =L$$ for your benefit).

Similarly there is an interval of type $$(b-h_2,b)$$ where the values of $$f$$ are less than $$k$$. Let $$p\in(a, a+h_1),q\in(b-h_2,b)$$ such that $$a. Then we have $$f(p) k> f(q)$$ and hence by applying intermediate value theorem in $$[p, d]$$ and $$[d, q]$$ we find $$a_1\in(p,d),b_2\in(d,q)$$ such that $$f(a_1)=f(b_1)=k$$. And now apply Rolle's theorem on $$f$$ on $$[a_1,b_1]$$.

The case when $$L$$ is infinite can't be handled by defining $$g$$ as in first part of this answer and that's where the second part comes into picture and one can see that this proof works fine (we need to choose $$k$$ between $$f(d)$$ and $$\pm\infty$$).